Exam (elaborations) TEST BANK FOR Adaptive Filter Theory 4th Edition By Simon Haykin (Solution manual only)

1.1 Let (1) (2) We are given that (3) Hence, substituting Eq. (3) into (2), and then using Eq. (1), we get 1.2 We know that the correlation matrix R is Hermitian; that is Given that the inverse matrix R-1 exists, we may write where I is the identity matrix. Taking the Hermitian transpose of both sides: Hence, That is, the inverse matrix R-1 is Hermitian. 1.3 For the case of a two-by-two matrix, we may ru(k) = E[u(n)u*(n – k)] ry(k) = E[y(n)y*(n – k)] y(n) = u(n + a) – u(n – a) ry(k) = E[(u(n + a) – u(n – a))(u*(n + a – k) – u*(n – a – k))] = 2ru(k) – ru(2a + k) – ru(– 2a + k) RH = R R –1RH = I RR –H = I R –H R –1 = Ru = Rs + Rν 2 For Ru to be nonsingular, we require With r12 = r21 for real data, this condition reduces to Since this is quadratic in , we may impose the following condition on for nonsingularity of Ru: where 1.4 We are given This matrix is positive definite because r11 r12 r21 r22 σ2 0 0 σ2 = + r11 σ2 + r12 r21 r22 σ2 + = det(Ru) r11 σ2 ( + ) r22 σ2 = ( + ) – r12r21 0 r11 σ2 ( + ) r22 σ2 ( + ) – r12r21 0 σ2 σ2 σ2 1 2 --(r11 + r22) 1 4Δr (r11 + r22)2 – 1 – --------------------------------------       Δr r11r22 r12 2 = – R 1 1 1 1 = aTRa [a1,a2] 1 1 1 1 a1 a2 = a1 2 2a1a2 a2